Module 0 Week 3 Challenge Q20
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I read through the solutions for Q20 and it makes perfect sense, but what I don't understand is why you can't solve this problem this way: list out all the possibilities of exiting (I counted 7 ways: 2 for A, 2 for B, 1 for C, D and E) and then circle out the ways that you exit from either A, B or C (5 ways) and compose that into the fraction 5/7. Could anyone help explain why?
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@fantasticcrow That's a really great question!
Listing out all the "exit possibilities" could definitely work-- but you have to be careful, because not all the possibilities are equally likely. For example, it's a lot more likely to go from the start point, to the top left room, and then out through E (probability 1/3) than it is to go from the start point, down, down, left, and out through C (probability 1/12, as calculated in the solution). This is mainly because the second path is "longer."
So basically, when you list out all possible ways to exit, not all the paths are going to be equally likely to occur. That's why you can't just count the number of paths out of A, B, and C and divide by the number of paths out from anywhere. (To put it a different way, we can't treat it like the probability of choosing a green marble if we have 5 green marbles and 7 total marbles-- the marbles are different sizes, which implies that some are more likely to be chosen than others)
Hope this helped!
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@audrey ooh, thank you so much! that was really helpful and i do get it now.
if you don't mind me asking, is there any way to determine what problems should use probability, and in which problems would counting possibilities work? thanks!!! -
@fantasticcrow Yay, I'm glad that helped!! And as for counting versus probability, it all depends on what the question asked for. To give a simple example:
- How many ways are there to make a three digit number where all the digits are even? [This would be a counting problem, because you want to count how many 3-dig #'s there are where all the digits are even]
Vs.
- If you choose any three digit number at random, what is the probability that all its digits are even? [This would be a probability problem, because you're asked what the chance of something happening is-- so in this case, you would take the answer from #1, and divide by how many three digit numbers there are!]
In the context of this problem, we can't use simple counting because not all of the options are equally likely to be chosen. In other words, you should use counting when all the options are equally likely ("weighted")-- for example, choosing a three digit number at random. But in this case, they're not, so we have to actually multiply out probabilities, if that made sense.
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@audrey said in Module 0 Week 3 Challenge Q20:
@fantasticcrow Yay, I'm glad that helped!! And as for counting versus probability, it all depends on what the question asked for. To give a simple example:
- How many ways are there to make a three digit number where all the digits are even? [This would be a counting problem, because you want to count how many 3-dig #'s there are where all the digits are even]
Vs.
- If you choose any three digit number at random, what is the probability that all its digits are even? [This would be a probability problem, because you're asked what the chance of something happening is-- so in this case, you would take the answer from #1, and divide by how many three digit numbers there are!]
In the context of this problem, we can't use simple counting because not all of the options are equally likely to be chosen. In other words, you should use counting when all the options are equally likely ("weighted")-- for example, choosing a three digit number at random. But in this case, they're not, so we have to actually multiply out probabilities, if that made sense.
Very helpful. Thank you!
